Question

78. In Fig. 9.32, area of ∆ AFB is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. What is the area of ∆DAO, where O is the mid point of DC?
Fig. 9.32

Answer 1

${\purple{\underline{\underline{\large{\mathtt{Answer:}}}}}}$

Given:

• We have been given that area of ∆AFB is equal to the area of parallelogram ABCD.
• The measure of altitude EF is 16cm.
• The measure of base AB is 10cm.
• O is the mid point of DC.

To Find:

• We need to find the altitude of the parallelogram to the base AB of length 10cm.
• We also need to find the area of ∆DAO, where O is the mid point of DC.

Solution:

We know that the area of triangle is given by 1/2 × base × altitude.

We also know that the area of parallelogram is base × height.

Now, it is given that area of ∆AFB is equal to the area of parallelogram ABCD.

=> 1/2 × AB × EF = DC × h

=> 1/2 × 10 × 16 = 10 × h

=> 1/2 × 160 = 10h

=> 80 = 10h

=> 80/10 = h

=> 8 = h

Therefore, h = 8cm.

Also, O is the midpoint of DC, therefore

DO = 1/2 × DC

=> DC = 1/2 × 10

=> DC = 5cm

Now, area of ∆DAO = 1/2 × DO × h

= 1/2 × 5 × 8

= 1/2 × 40

= 20cm^2.